Why can we multiply by breaking up the factors as sums in different ways? My friend and I were discussing some mathematical philosophy and how the number systems were created when we reached a question. Why can we multiply two different numbers like this?
Say we had to multiply $13\times 34$. One may break this up like $(10+3)\times (30+4)$. Applying distributive property here will give us the answer of 442. We can also choose to multiply this as $(6+7)\times (22+12)$. Intuitively, we can hypothesize that this should give us the same answer as $13\times 34$. How can we prove that our answer will be equal regardless of how we break up the numbers?
 A: The simplest and historical answer is that we observed that certain things in the world can be counted, in the sense that we can put them in a line and label each one with their position in the line, and that these positions satisfy certain empirical facts, and hence we invented axioms to capture these facts. I specifically single out the axioms in that linked section (discrete ordered semi-ring axioms plus induction), rather than the other more commonly known axiomatization of PA, for two reasons.
Firstly, the ring axioms have been observed or proven to hold for a very wide variety of structures, so it is no surprise that people invent them, and they include distributivity of multiplication over addition.
Secondly, that specific axiomatization I linked to accurately captures our empirical knowledge of counting numbers. For example, from our understanding of spatial motions (translations and rotations) we readily grasp that addition is commutative: $a+b$ is the total count of $a$ objects followed by $b$ objects in one line, which is the same as $b+a$ by looking at the line from the other side. Similarly for the commutativity of multiplication: $a·b$ is the total count of a rectangular array of $a$ times $b$ objects, which is equal to $b·a$ by rotating the array by $90$ degrees. And distributivity ($a·(b+c) = a·b+a·c$) is clear by splitting the array. Associativity is also intuitive.
Note that the above axioms are sufficient for us to recover general distributivity of the form you are using: Observe that $(a+b)·c = c·(a+b) = c·a+c·b = a·c+b·c$ by commutativity and distributivity. Thus $(a+b)·(c+d) = (a+b)·c + (a+b)·d = (a·c+b·c) + (a·d+b·d)$, and the order of addition in the last expression does not matter by associativity.

There is a more complicated answer, that also goes a significant way to explaining why counting numbers are the way they are. If you have an operator $f$ on some collection $S$ that can be iterated (i.e. $f : S \to S$), then you can define $f^0 = \text{id}_S$ and $f^1 = f$, and define $f^{m+n} = f^m \circ f^n$ and $f^{m·n} = (f^m)^n$ for every counting numbers $m,n$. The last definition is valid because (by induction) $f^m$ is an operator on $S$ that can be iterated. Then (again by induction), you can actually prove the basic properties of $+,·$. For example:
Associativity of $+$
Take any naturals $k,m,n$. Then $f^{k+(m+n)} = f^k \circ ( f^m \circ f^n ) = ( f^k \circ f^m ) \circ f^n = f^{(k+m)+n}$. Here we are using the fact that function composition is associative, which is a core reason for almost every instance of associativity in mathematics.
Commutativity of $+$
First we show that $f^{m+1} = f^{1+m}$ for every natural $m$. $f^{0+1} = f^0 \circ f = f = f$ $\circ f^0 = f^{1+0}$ by definition. For any natural $n$ such that $f^{n+1} = f^{1+n}$, we also have $f^{(n+1)+1} = f^{n+1} \circ f = f^{1+n} \circ f$ $= f^{(1+n)+1} = f^{1+(n+1)}$. Therefore by induction we are done.
Now take any natural $m$. Then $f^{m+0} = f^m \circ f^0 = f^m$ $= f^0 \circ f^m = f^{0+m}$. And for any natural $n$ such that $f^{m+n} = f^{n+m}$, we also have $f^{m+(n+1)} = f^{(m+n)+1} = f^{m+n} \circ f = f^{n+m} \circ f$ $= f^{(n+m)+1} = f^{n+(m+1)}$ $= f^n \circ f^{m+1} = f^n \circ f^{1+m}$ $= f^{n+(1+m)} = f^{(n+1)+m}$. Therefore by induction we are done.
As you can see from the above proofs, induction is enough for us to bootstrap from very rudimentary notions of iteration to obtain addition and its properties. We can do the same for multiplication. But it should be clear that this was not how the axioms were originally invented.
A: The "reason" it works is because in this universe quantities stay the same no matter how we group or arrange them.  Therefore because things don't magically appear and disappear we can count them.  And if we wanted to combine things by counting one group and then another group and combing them with another we can add them.  
Thus we no $10 + 3 = 7+ 6 = 13$ are consistent ways of grouping and identifying values.
And we can do multiple adding to define multiplication. So $6 +6 + 6+6 =6\times 4 = 24$.  And as it quantities don't change when we group them then  $n(a + b) = \underbrace{(a+b)+... + (a+b)}_{n} = \underbrace{a + a...+a}_{n} + \underbrace{b + b...+b}_{n}=n\times a + n\times b$.
That is the way the world works.
And that is what we teach children.
But what kind of unimaginative simpleton cares about how the real world works?  Certainly not mathematicians.
Math is modeling and systemizing.
There are two standards:
1)  Axiomatic definitions.  The definition is a field, $F$ includes the axiom that for any $a,b,c \in F$ that $a(b + c) = ab + ac$.  That is the algebrists saying "I don't care why this is true in the real world, but it is this way in my world because I say it is."
2) Construction:  Develop the concept of natural numbers via the Peano Postulates and defining the idea of a first element $0$ and the ability to find a unique successor and a few basic axioms.  ... And then prove distribution.  In essence it is the same as how the real works, except we have defined the concept of number purely and abstractly, and not by giving Fred a bunch of apples.
This is the constructionist way of saying "I've extracted the pure essence of the real world into consistent abstract concepts".
A: There are two ways I see to answer this question. One is from an axiomatic standpoint, where numbers are merely symbols on paper that are required to follow certain rules. The other uses the interpretation of multiplication as computing area. The former would take a good 5-10 pages to build up from the Peano axioms.
For the latter, you can draw a rectangle 13 units by 34 units. Break one side into 10 units and 3 units, and the other side into 30 units and 4 units. At this point, you should see that this decomposes the rectangle into four pieces, corresponding to the four terms you get from the distributive rule. The various was to compute $13 \cdot 34$ are all just ways to decompose the rectangle, and at the end of the day they all compute the same number: the area of the rectangle. 
A: Well, that is literally what the distributive law tells you.  It tells you that $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$ and so whenever you break up the two factors of a product as a sum, you can use the "pieces" to compute the product.
So what you are really asking for is a proof of the distributive law itself.  What constitutes a "proof" of such a basic fact depends heavily on what definitions of "numbers" and the operations on them that you are using (for some definitions, the distributive law is simply an axiom that you assume).  But here is an intuitive explanation that works for natural numbers (and this can be turned into a rigorous proof if you define arithmetic of natural numbers in terms of cardinalities of sets).
We want to prove that $(a+b)c=ac+bc$.  What does a product $xy$ of natural numbers mean?  Well, it means you draw a grid of dots with $x$ rows and $y$ columns, and count up the total number of dots.  So, to compute $(a+b)c$, you draw a grid with $a+b$ rows and $c$ columns.  Now we observe that we can split such a grid into two pieces: the top $a$ rows and the bottom $b$ rows.  The top $a$ rows form a grid with $a$ rows and $c$ columns, so they have $ac$ dots.  The bottom $b$ rows form a grid with $b$ rows and $c$ columns, so they have $bc$ dots.  In total, then, we have $ac+bc$ dots, so $(a+b)c=ac+bc$.
(In the computation of $(a+b)(c+d)$ above I used the distributive law in two different versions, one with the sum on the left side of the product and one with the sum on the right side of the product.  You can prove the version with the sum on the right side of the product in the same way (you just split up the columns instead of the rows), or you can deduce it from the other version using commutativity of multiplication.)
A: 
How can we prove that our answer will be equal regardless of how we break up the numbers?

I may be not getting the gist of your question, other answers seem deep; but on some (shallowish) level you can proceed as follows.
If you have $A$ and $B$ and you choose $a$ and $b$ freely, then
$$A \equiv a+(A-a)$$
$$B \equiv b+(B-b)$$
Note I use "$\equiv$" symbol that is stronger than "$=$"; see this answer of mine (its part named "definitions vs. equations or conditions") to get the difference. In the above expressions I used "$\equiv$" to emphasize they are true regardless of $a$ and $b$ (compare $A=A+b$ which may or may not be true (depending on $b$), hence only "$=$", not "$\equiv$").
In your example $A=13$ and $B=34$; initially you choose $a=10$ and $b=30$ and you calculate $(A-a)=3$ and $B-b=4$. Whatever numbers you choose for $a$ and $b$, your answer is expressed as
$$W(A,B,a,b) \equiv (a+(A-a))\times(b+(B-b)) \tag{*}$$
Now use
$$x-y \equiv x+(-1)\times y$$ to turn subtraction into addition; and use distributive property to get rid of the parentheses one by one. You don't want to cancel out $a$ and $-a$ right away, this would be trivial (as if you didn't break up the number); instead apply distributive property first in the same way you did with your numbers, observe that some expressions cancel out eventually. The outcome is:
$$W(A,B,a,b) \equiv A \times B \tag{**}$$
I don't know if this is what you asked for. Still on some level it shows that the answer (i.e. $W$) for fixed $A$ and $B$ is equal (i.e. the same) regardless of how we break up the numbers (i.e. regardless of what $a$ and $b$ we choose).

Formally for all this to work $\times$ and $+$ must denote some operations (with certain properties) that you can perform on members of some set (a set $A$, $B$, $a$, $b$, $0$, $1$ and $-1$ must belong to). It all comes to the concept of a ring this answer has already introduced. It's a ring with unity. The concept requires $0$, $1$ and $-1$ to be somewhat special:
$$0+x \equiv x$$
$$1 \times x \equiv x$$
$$1+(-1)=0$$
Plus (in our case) there is commutativity of $+$ and $\times$. If you go from (*) to (**) step by step and in a formal way then you will notice you need to use almost all these properties.
A: Formally, this is a property of rings.  Rings have a multiplication operation that distributes with respect to addition, meaning for any 3 numbers $a$ $b$ and $c$: $$a\cdot(b + c) = (a\cdot b) + (a\cdot c)$$ $$(b + c)\cdot a = (b\cdot a) + (c\cdot a)$$  The real numbers are a ring (they're actually a field, which is a special kind of ring), so that's half of your answer.
The other half is Eric's answer which addresses the natural numbers.  Natural numbers are not a ring, but they do have a distributive property.
From a philosophical perspective, we could define anything we wanted, but what's interesting about this particular pattern is that it's so useful.  Nothing prevents me from making $x+\frac{3}{2}$ means a triple gainer with a twist, but outside of diving, that particular pattern isn't all that useful.  We tend to find that fields and rings show up rather often in physically meaningful scenarios.
Now from a philosophical perspective, it makes sense to point out that there are also lots of other really useful patterns that show up.  For example, if you look at how we define rotations, such as using yaw, pitch, and roll to describe the orientation of an aircraft, those don't seem to add the way we want them to.  The rotations form a pattern known as a group, which doesn't even have a concept of addition at all!  They only have multiplication.  And by that I mean mathematicians decided to call the one operation in this pattern "multiplication" because its rules are a generalization of matrix multiplication.
We also have all sorts of oddball cases which may or may not actually be philosophically relevant.  For example, we can consider the ordinal numbers, which explore labeling objects as 1st, 2nd, 3rd, and so on.  Ordinals grapple with the concept of infinity, which generally means they've got some quirks.  One of the quirks of ordinals is that they are left distributive but not right distributive.  That means I can use the distributive property in $a\cdot(b + c)$ but not $(b + c)\cdot a$!  So that shows that we've come up with some really strange systems which look sane, but where the distributive law starts to get a little strange.  (For what it's worth, part of the reason this law acts so strange is that multiplication isn't commutative in ordinals: $2\cdot\omega \neq \omega\cdot 2$)
So in the end, what makes this distributive law so interesting philosophically is that real numbers and natural numbers seem to be terribly good at describing the world around us, and both of them have distributive properties.  But that doesn't mean that everything interesting has a distributive law, or even that the distributive law will make intuitive sense to you!  Now the question for why real numbers and natural numbers are so useful in reality is a really interesting philosophical question which has lead some people to argue that mathematics is the language upon which reality sits.
I say it sits on a turtle.  But who am I to judge?
A: For any natural numbers $a$ and $b$ and for any two ordered quadruplets of natural numbers (c, d, e, f) and (g, h, i, j) where $a = c + d$; $b = e + f$; $a = g + h$ and $b = i + j$, $(c + d)(e + f) = (g + h)(i + j)$ can be proven as follows.
$(c + d)(e + f) = ab = (g + h)(i + j)$
This question can also be interpreted as follows: how can I prove that $ce + ef + de + df = gi + gj + hi + hj?$
This is how to do it. It is a theorem that for any natural numbers $k$, $l$, $m$, $n$, $(k + l)(m + n) = km + kn + lm + ln$ so
$ce + cf + de + df = (c + d)(e + f) = ab = (g + h)(i + j) = gi + gj + hi + hj$
